# Cannot solve equation

I just started using mathematica and I'm facing a problem that I just can't solve. I want to solve the following equation:

Code:

eqn = {(ao)/((c*Exp[a*(n - o)/n]) (-Log[(c*Exp[a*(n - o)/n])] + Log[c] + a)^2) == m,
ao*1/2 < n};

Solve[eqn, n, Reals]


Program says ' This system cannot be solved with the methods available to Solve.'

Can anybody help me with this?

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This equation is highly nonlinear equation. you may use FindRoot but you need to provide numerical values to other variables. –  Algohi Jun 24 '14 at 15:06
Thank you very much for your quick reply. I tried using FindRoot but didn't succeed. –  Andy Jun 24 '14 at 15:39
You may start by something like this r = Reduce[eqn[[1]] /. a*(n - o)/n -> x /. {n -> (a o)/(a - x)}, x, Reals]; FullSimplify[r /. x -> a*(n - o)/n, Assumptions -> ao*1/2 < n] –  belisarius Jun 24 '14 at 15:44
If you give specific values for the parameters ao, a, etc. then it becomes amenable to Solve. –  Daniel Lichtblau Jun 24 '14 at 15:57
@Daniel: Is that so?! Thx, I will try that. –  Andy Jun 24 '14 at 16:11

If you use FindRoot you can get some solutions. I don't know what range of values that you are expecting but I will assume some.

eq = (ao)/((c*
Exp[a*(n - o)/n]) (-Log[(c*Exp[a*(n - o)/n])] + Log[c] + a)^2)
Plot[{eq /. {a -> 1, c -> 1, o -> 1, ao -> 1}, 20}, {n, 0, 10}]


for this set of values you will have two roots. you can find them as follows:

    FindRoot[(eq /. {a -> 1, c -> 1, o -> 1, ao -> 1}) == 20, {n, 2}]
(* {n -> 6.85462} *)

FindRoot[(eq /. {a -> 1, c -> 1, o -> 1, ao -> 1}) == 20, {n, .2}]
(* {n -> 0.121873} *)

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Looks good! Although, I was hoping to create a universal solution without using values. –  Andy Jun 24 '14 at 16:13
This is why i added the term (ao)/2 < n. This is the inflection point, so I'm just looking on the right part of the function with only one solution for n. –  Andy Jun 24 '14 at 16:17
@Andy as I said, the equation is highly nonlinear with multi variables. I don't think you can find a universal solution to the problem. you can get bunch of solutions and then use regression analysis to find empirical relationship. –  Algohi Jun 24 '14 at 16:32